Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))
f'3(s1(x0), x1, x1)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F1(g1(x)) -> F1(f1(x))
F'3(s1(x), y, y) -> F'3(y, x, s1(x))
F1(g1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))
f'3(s1(x0), x1, x1)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F1(g1(x)) -> F1(f1(x))
F'3(s1(x), y, y) -> F'3(y, x, s1(x))
F1(g1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))
f'3(s1(x0), x1, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F1(g1(x)) -> F1(f1(x))
F1(g1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))
f'3(s1(x0), x1, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F1(g1(x)) -> F1(f1(x))
F1(g1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  x1
g1(x1)  =  g1(x1)
f1(x1)  =  x1
h1(x1)  =  h

Lexicographic Path Order [19].
Precedence:
h > g1


The following usable rules [14] were oriented:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f1(g1(x)) -> g1(f1(f1(x)))
f1(h1(x)) -> h1(g1(x))
f'3(s1(x), y, y) -> f'3(y, x, s1(x))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))
f'3(s1(x0), x1, x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.